nZn = 13/65 = 0,2 (mol)

mHCl = 182,5 . 10% = 18,25 (g)

nHCl = 18,25/36,5 = 0,25 (mol)

PTHH: Zn + 2HCl -> ZnCl2 + H2

LTL: 0,2 < 0,5/2 => HCl dư

nH2 = nZnCl2 = nZn = 0,2 (mol)

VH2 = 0,2 . 22,4 = 4,48 (l)

mH2 = 0,2 . 2 = 0,4 (g)

Chất tan: ZnCl2 và HCl dư

nHCl (p/ư) = 0,2 . 2 = 0,4 (mol)

mHCl (dư) = (0,5 - 0,4) . 36,5 = 3,65 (g)

mZnCl2 = 136 . 0,2 = 27,2 (g)

mdd (sau p/ư) = 182,5 + 13 - 0,4 = 195,1 (g)

C%HCl = 3,65/195,1 = 1,87%

C%ZnCl2 = 27,2/195,1 = 13,94%

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{ddHCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ C\%_{ddZnCl_2}=\dfrac{136.0,2}{13+146-0,2.2}.100\approx17,15\%\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:    0,2     0,4         0,2       0,2

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

\(m_{ddHCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\)

m_{dd sau pứ} = 13+146-0,2.2 = 158,6 (g)

\(C\%_{ddZnCl_2}=\dfrac{0,2.136.100\%}{158,6}=17,15\%\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
\(pthh:Zn+2HCl->ZnCl_2+H_2\)
 LTL :
\(\dfrac{0,2}{1}=\dfrac{0,4}{2}\)
=> ko chất nào dư 
theo pthh : \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\\ =>V_{H_2}=0,2.22,4=4,48\left(l\right)\)
 

Bài 8 (bài 7 bạn ở trên làm rồi)

\(n_{H_2\left(Al\right)}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{H_2\left(Zn\right)}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH:

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

0,2      0,3                0,1            0,3

Zn + H₂SO₄ ---> ZnSO₄ + H₂

0,3     0,3              0,3        0,3

\(a,\left\{{}\begin{matrix}m_{Al}=0,3.27=8,1\left(g\right)\\m_{Zn}=0,3.65=19,5\left(g\right)\end{matrix}\right.\\ b,m_{H_2SO_4}=\left(0,3+0,3\right).98=58,8\left(g\right)\)

c, Hợp chất tạo thành thuộc loại muối trung hoà

m_muôí = 0,1.342 + 0,3.161 = 82,5 (g)

\(pthh:\left\{{}\begin{matrix}Al+3H_2SO_4->Al_2\left(SO_{\text{4}}\right)_3+3H_2\left(1\right)\\Zn+H_2SO_4->ZnSO_4+H_2\left(2\right)\end{matrix}\right.\)
\(n_{H_2\left(1\right)}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{H_2\left(2\right)}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
=>\(m_{Al}=0,3.27=8,1\left(mol\right)\\ n_{Zn}=0,2.65=13\left(g\right)\)
\(\Sigma n_{H_2SO_4}=0,3+0,2=0,5\left(mol\right)\\ =>\Sigma m_{H_2SO_4}=0,5.98=49\left(g\right)\)
=>\(\left\{{}\begin{matrix}m_{Al_2\left(SO\right)_4}=342.0,1=34,2\left(G\right)\left(mu\text{ối}\right)\\m_{ZnSO_4}=0,2.161=32,2\left(g\right)\left(mu\text{ối}\right)\end{matrix}\right.\)

\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

\(n_{HCl}=\dfrac{182.5\cdot10}{100\cdot36.5}=0.5\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.2......0.4..........0.2........0.2\)

\(n_{HCl\left(dư\right)}=0.5-0.4=0.1\left(mol\right)\)

\(m_{HCl\left(dư\right)}=0.1\cdot36.5=3.65\left(g\right)\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{\text{dung dịch sau phản ứng}}=13+182.5-0.2\cdot2=195.1\left(g\right)\)

\(C\%_{HCl\left(dư\right)}=\dfrac{3.65}{195.1}\cdot100\%=1.87\%\)

\(C\%_{ZnCl_2}=\dfrac{0.2\cdot136}{195.1}\cdot100\%=13.94\%\)

lập pthh của pư

zn+2hcl→zncl2 +h2

1mo  2mol   1mol  1mol

0,2mol           0,2mol       0,2mol 

số mol zn 

nzn=\(\frac{13}{65}\)=0,2 mol

thể tích khí H2 

VH2 = 0,2 . 22,4 =4,48 lít 

khối lượng zncl2 

mzncl2= 0,2.  136=2,72 gam 

nồng độ % dd zncl2

c% zncl2=\(\frac{mct}{mdd}\). 100%=\(\frac{2,72}{200}\). 100%=1,36%

$n_{Zn} = \dfrac{6,5}{65} = 0,1(mol) \\ PTHH: Zn + 2HCl \to ZnCl_2 + H_2 \\$$n_{H_2} = n_{Zn} = 0,1(Mol) \\ V_{H_2} = 0,1.22,4 = 2,24l \\b) PTHH: H_2 + CuO \xrightarrow[]{t^o} Cu + H_2O \\ n_{CuO} = \dfrac{12}{64} = 0,15(mol) \\ \to CuO dư$ $\\ n_{H_2} = n_{Cu} = 0,1(mol \\ m_{Cu} = 0,1.64 = 6,4(gam)$

Zn+2HCl->ZnCl2+H2

0,1---0,2----0,1---0,1

n Zn=0,1 mol

=>VH2=0,1.22,4=2,24l

H2+CuO-to>Cu+H2O

          0,15-----0,15

n CuO=0,15 mol

=>H2 dư

=>m Cu=0,15.64=9,6g

Bài 1:

a) nP=6,2/31=0,2(mol); nO2= 6,72/22,4=0,3(mol)

PTHH: 4P + 5 O2 -to-> 2 P2O5

Ta có: 0,2/4 < 0,3/5

=> P hết, O2 dư, tính theo nP

=> nO2(p.ứ)= 5/4. nP= 5/4. 0,2=0,25(mol)

=> mO2(dư)=0,3- 0,25=0,05(mol)

=> mO2(dư)=0,05.32=1,6(g)

b) nP2O5= nP/2= 0,2/2=0,1(mol)

=>mP2O5=0,1.142=14,2(g)

Bài dài quá b ạ 

b ơi đây là bài 1 nka, tí mk làm xog sẽ gửi típ nka 

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2(mol)\)

Vì \(\dfrac{n_{Zn}}{1}>\dfrac{n_{HCl}}{2}\) nên Zn dư

\(\Rightarrow n_{Zn({\text{phản ứng})}}=\dfrac{1}{2}n_{HCl}=0,15(mol)\\ \Rightarrow n_{Zn(\text{dư})}=0,2-0,15=0,05(mol)\\ \Rightarrow m_{Zn(\text{dư})}=0,05.65=3,25(g)\\ c,n_{ZnCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,15(mol)\\ \Rightarrow a=m_{ZnCl_2}=0,15.136=20,4(g)\\ V=V_{H_2}=0,15.22,4=3,36(l)\)

3,36

\(a,Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ Ta.c\text{ó}:\dfrac{0,2}{1}>\dfrac{0,3}{2}\Rightarrow Zn.d\text{ư}\\ b,n_{Zn\left(d\text{ư}\right)}=0,2-\dfrac{0,3}{2}=0,05\left(mol\right)\\ \Rightarrow m_{Zn\left(d\text{ư}\right)}=0,05.65=3,25\left(g\right)\\ c,a=m_{ZnCl_2}=0,15.136=20,4\left(g\right)\\ V=V_{H_2\left(\text{đ}ktc\right)}=0,15.22,4=3,36\left(l\right)\)

nZn = 13/65 = 0,2 (mol)

nHCl = 14,6/36,5 = 0,4 (mol)

PTHH: Zn + 2HCl -> ZnCl2 + H2

LTL: 0,2 = 0,4/2 => phản ứng đủ

nH2 = nZnCl2 = 0,2 (mol)

VH2 = 0,2 . 22,4 = 4,48 (l)

mZnCl2 = 0,1 . 136 = 27,2 (g)

mH2 = 0,2 . 2 = 0,4 (g)

\(n_{Zn}=\frac{13}{65}=0,2\left(mol\right)\)

\(n_{HCl}=\frac{14,6}{36,5}=0,4\left(mol\right)\)

\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

tỉ số: \(\frac{0,2}{1}=\frac{0,4}{2}\Rightarrow\) PTHH này ko có chất dư (sau khi pứ kết thúc)

a) \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\)

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

\(n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\)

\(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

XONG!